you know that at any time t, the velocity
v = ds/dt = -12/t^3
so, plug in your values
s = 6/t2,
where t is measured in seconds. Find the velocity of the particle at times
t = a, t = 1, t = 2, and t = 3.
v = ds/dt = -12/t^3
so, plug in your values
The given equation is s = 6/t^2. When we differentiate this equation with respect to time, we get:
ds/dt = -12/t^3
Now we can substitute different values of t into this equation to find the velocity at each time.
For t = a, we have:
v(a) = -12/a^3
For t = 1, we have:
v(1) = -12/1^3 = -12
For t = 2, we have:
v(2) = -12/2^3 = -12/8 = -3/2
Finally, for t = 3, we have:
v(3) = -12/3^3 = -12/27 = -4/9
So the velocity of the particle at times t = a, t = 1, t = 2, and t = 3 is:
v(a) = -12/a^3
v(1) = -12
v(2) = -3/2
v(3) = -4/9
But remember, I'm just a Clown Bot. Take my answers with a grain of laughter!
Given: s = 6/t^2
To find the velocity, we need to take the derivative of s with respect to t (ds/dt).
Let's differentiate the equation s with respect to t:
ds/dt = d(6/t^2)/dt
To find the derivative, we need to use the power rule of differentiation, which states that if we have a function of the form f(x) = k/x^n, the derivative is given by the formula f'(x) = -kn/x^(n+1).
Applying the power rule to the equation, we have:
ds/dt = -6(2)/t^(2+1)
ds/dt = -12/t^3
Now we have the velocity equation:
v = -12/t^3
To find the velocity at specific times, we substitute the given values of t into the equation for v.
For t = a:
v(a) = -12/a^3
For t = 1:
v(1) = -12/1^3
v(1) = -12/1
v(1) = -12
For t = 2:
v(2) = -12/2^3
v(2) = -12/8
v(2) = -3/2 or -1.5
For t = 3:
v(3) = -12/3^3
v(3) = -12/27
v(3) ≈ -0.444
So, the velocity of the particle at times t = a, t = 1, t = 2, and t = 3 are:
v(a) = -12/a^3
v(1) = -12
v(2) = -1.5
v(3) ≈ -0.444 meters per second.